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Daily CSAT Practice Test
Everyday 5 Questions from Aptitude, Logical Reasoning, and Reading Comprehension will be covered from Monday to Saturday.
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A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
Solution (d)
We have: l = 20 ft. and lb = 680 sq ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft. = 88 ft.
Solution (d)
We have: l = 20 ft. and lb = 680 sq ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft. = 88 ft.
Prema is twice as efficient as Anjali and takes 90 days less than Anjali to complete the job. Find the time in which they can finish the job together?
Solution (d)
Since ‘A’ is ‘m’ times as efficient as ‘B’ & takes ‘D’ days less than ‘B’, then the time required to complete the job together is given by,
T = m × D / (m2-1)
T = 2 × 90 / (4-1)
T = 60 days
Solution (d)
Since ‘A’ is ‘m’ times as efficient as ‘B’ & takes ‘D’ days less than ‘B’, then the time required to complete the job together is given by,
T = m × D / (m2-1)
T = 2 × 90 / (4-1)
T = 60 days
A sales executive went to meet his client A in another office 17 km away in the North-east direction of his own office. From there he went to see client B in an office 15 km in the south of his client A’s office. How far from the starting point and in what direction is he now?
Solution (b)
Solution (b)
Given below are the statement and 2 conclusions. Based on that select the correct option
Statement: T ≤ M ≤ J = Q > F > S; M > R
Conclusions:
Choose the correct code
Solution (c)
T ≤ M ≤ J = Q > F > Sà(1)
M > R à(2)
From (1), we get
T ≤ Q or Q ≥ T. Thus, conclusion I is true
Again, J > S is true. Hence both conclusion I and II are true.
Solution (c)
T ≤ M ≤ J = Q > F > Sà(1)
M > R à(2)
From (1), we get
T ≤ Q or Q ≥ T. Thus, conclusion I is true
Again, J > S is true. Hence both conclusion I and II are true.
Given below are the statement and 2 conclusions. Based on that select the correct option
Statement: H > S ≥ F = B ≤ U ≤ T; E ≤ B ≤ K
Conclusions:
Choose the correct code
Solution (a)
H > S ≥ F = B ≤ U ≤ Tà(1)
E ≤ B ≤ Kà(2)
Combining (1) and (2), we get
E ≤ B ≤ U
Thus, E ≤ U or U ≥ E. Hence I is true. Again we can’t compare S and T. Hence II (S > T) is not true.
Solution (a)
H > S ≥ F = B ≤ U ≤ Tà(1)
E ≤ B ≤ Kà(2)
Combining (1) and (2), we get
E ≤ B ≤ U
Thus, E ≤ U or U ≥ E. Hence I is true. Again we can’t compare S and T. Hence II (S > T) is not true.
